This is an expansion and follow up of my comment.
As mentioned in comment.
$\hspace0.5in$ Without further restriction, there are no relation.
An example is the function $f(x) = x^3$ which is invertible over the real axis and yet its inverse function doesn't have a power series expansion at $x = 0$.
In general, if your function is invertible only over real axis (or part of it), there isn't much one can do but check the power expansions of the function and its inverse function separately.
However, if your function is analytic and injective over some open subset of $\mathbb{C}$ (such a function is known as an univalent function), there is a theorem by Koebe which can help you.
Koebe quarter theorem
The image of any univalent function $\varphi : B(0,1) \to \mathbb{C}$ from the unit disk $B(0,1)$ onto a subset of the complex plane contains the disk $\displaystyle\;B\left( \varphi(0), \frac{|\varphi'(0)|}{4}\right)\;$.
Let's say your function $f$ is univalent over the open disk $B(z_0,R)$. Since $f$ is analytic and locally injective at $z = z_0$, $f'(z_0) \ne 0$. Now define a function $\varphi : B(0,1) \to \mathbb{C}$ by
$$B(0,1) \ni \omega \mapsto \varphi(\omega) = f(z_0 + R \omega ) \in \mathbb{C}$$
We have $\varphi(0) = f(z_0)$ and $\varphi'(0) = R f'(z_0)$. By Koebe quarter theorem,
$$f(B(z_0,R)) = \varphi(B(0,1)) \supset B\left( f(z_0), \frac{R|f'(z_0)|}{4} \right)$$
This implies the nearest singularity of the inverse function $\varphi^{-1}$ in $\mathbb{C}$ is at least at a distance $\displaystyle\;\frac{R|f'(z_0)|}{4}$ from $f(z_0)$. As a result, the radius of convergence of the power series of $\varphi^{-1}$ and hence that of $f^{-1}$ at $f(z_0)$ is at least $\displaystyle\;\frac{R|f'(z_0)|}{4}$.
Your answer is correct. One may recall that
$$
\frac1{1+u}=\sum_{n=0}^\infty (-1)^n u^n, \quad |u|<1,
$$ giving, for $2x^2<1$,
$$
\frac1{1+2x^2}=\sum_{n=0}^\infty (-2)^n x^{2n}
$$ that is
$$
\frac{x}{1+2x^2}=\sum_{n=0}^\infty (-2)^n x^{2n+1}, \quad x \in \left(-\frac{\sqrt{2}}2,\frac{\sqrt{2}}2\right).
$$
Best Answer
Outline: Our function can be rewritten as $$f(x)=\frac{x^3}{2^3}\cdot \frac{1}{(1-x/2)^3}.$$ To find the expansion of $\frac{1}{(1-t)^3}$, note that for suitable $t$ we have $$\frac{1}{1-t}=1+t+t^2+t^3+\cdots.$$ Differentiate twice with respect to $t$.